# Math Help - Field Theory - Basic Theory - D&F Section 13.1

1. ## Field Theory - Basic Theory - D&F Section 13.1

I am reading Dummit and Foote (D&F) Section 13.1 Basic Theory of Field Extensions.

I have a question regarding the nature of extension fields.

Theorem 4 (D&F Section 13.1, page 513) states the following (see attachment):

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Theorem 4. Let $p(x) \in F[x]$ be an irreducible polynomial of degree n over a field F and let K be the field $F[x]/(p(x))$. Let $\theta = x \ mod \ (p(x)) \in K$. Then the elements

$1, \theta, {\theta}^2, ... ... , {\theta}^{n-1}$

are a basis for K as a vector space over F, so the degree of the extension is n i.e.

$[K \ : \ F] = n$. Hence

$K = \{ a_0 + a_1 \theta + a_2 {\theta}^2 + ... ... + a_{n-1} {\theta}^{n-1} \ | \ a_0, a_1, ... ... , a_{n-1} \in F \}$

consists of all polynomials of degree $\lt n$ in $\theta$

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However, when we come to Example 4 on page 515 of D&F we read the following: (see attachment)

(4) Let $F = \mathbb{Q}$ and $p(x) = x^3 - 2$ which is irreducible by Eisenstein.

Denoting a root of p(x) by $\theta$ we obtain the field

$\mathbb{Q}[x]/(x^3 - 2) \cong \{a + b \theta + c {\theta}^2 \ | \ a, b, c \in \mathbb{Q}$

with ${\theta}^3 = 2$ an extension of degree 3. ... ... etc

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Now my problem is that in Theorem 4 we read

$K = \{ a_0 + a_1 \theta + a_2 {\theta}^2 + ... ... + a_{n-1} {\theta}^{n-1} \ | \ a_0, a_1, ... ... , a_{n-1} \in F \}$ which becomes

$K = \{a + b \theta + c {\theta}^2$ in the situation of Example 4

But then in Example 4 we have

$K = \mathbb{Q}[x]/(x^3 - 2) \cong \{a + b \theta + c {\theta}^2 \ | \ a, b, c \in \mathbb{Q}$

???

It seems that in Theorem 4, we have $\theta = x \ mod \ (p(x))$ but in Example (4) we have $\theta = \sqrt[3]{2}$ and we do not have equality but only an isomorphism, that is $\mathbb{Q}[x]/(x^3 - 2) \cong \mathbb{Q}(\sqrt[3]{2}$.

In Field theory we seem to prove that an irreducible polynomial has a root in a field that is isomorphic to the actual field that contains the root.

Does what I am saying make sense? Can someone clarify this issue for me?

Peter